Simultaneously Stokes and anti-Stokes Raman amplification in silica fiber
Stokes Lecture
Transcript of Stokes Lecture
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3
q
f) Gausss Law
!
E =1
4 !" 0
q
r2 er
For a sphere centred on q ,
!
E i d !
! =
! 1
""1
4 #$ 0
qr 2 d !
! 1
"" =
1
4 !" 0qr 2 d
#
# 1
!$= q
" 0
For a point charge,
Flux through any surface ( 2) containing no charges,
! E i d
!! =
! 2
"" # i!
E d $ $ 2
" (divergence theorem)= 0 (continuity for const density and no sources)
! E i d
!! =
! 3
"" # i!
E d $ $ 3
"Flux through any surface ( 3) containing charge q,
= ! i!
E d "
" 3#$ + ! i
!
E d "
" 3####$ = q
! 0
! E i d
!! =
!
"" q inside# 0
Gausss Law:
3 3
q
1
Since q inside = ! d " "
# where = source (charge) density, and
! E i d
!! =
!
"" # i!
E d $ $
"
then
! i!
E d " "
# = $ % 0
d " "
#
so
! i!
E ="
# 0differential form of Gausss Law
11) Stokes theorema) 2d
Greens theorem:
! Q! x
" ! P! y
# $ %
& '(
A
)) dxdy = (P dx + Q dy )! A
!)
For !
V = P
i + Q
j + R
k ,
(! "!
V ) i k =#Q# x
$ #P# y
% & '
( )*
and for d !
r = dx
i + dy
j , !
V i d !
r = Pdx + Qdy
so
! "!
V ( ) A
## i kdxdy = (!
V i d !r )
$ A"#
This is special case of Stokes theorem:
! "!
V ( )#
$$ i nd # = (!
V i d !r )
%# "$
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b) Interpretation of ! "!
V !!
!r
!
V
Recall!
V =!
! "!
r
Consider a uid rotating about the z-axis
At any point P,!
V =!
! "!
r
Then, ! "!
V = ! " (!
# "!
r )
P
=
!
! (" i!
r ) # (!
! i " )!
r
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! "
!
V =!
# (! i!
r ) $ (!
# i ! )!
r
but
! i!
r =" x
" x+
" y
" y+
" z
" z= 3
and
(!
! i " )!
r = ! x#
# x xi + ! y
#
# y yi + ! z
#
# z zi =
!
!
so ! "!
V = 2!
# is a measure of uid rotation -- hence curl.
For an arbitrary velocity eld, the curl represents the microscopiccirculation:
Place a sphere at a point in the uid, and its rotation represents the curl
If the sphere is on an axis, the rotation gives the component of the curl on thataxis.
If the curl is zero, the eld i s irrotational at that point.
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c) Line integral as circulation
Circulation of V :!
V i d !r"!
From Greens theorem in the xy plane,
(!
V i d !r )! A"" = # $
!V ( )
A"" i kdA
Example: !
V ( x, y) = yi ! x j
! "!
V =
i j k #
# x
#
# y
#
# z
y $ x 0
= k ($ 1 $ 1) = $ 2 k
(uniform)
!V i d
!r
C
"! = ( ydx " x dy )C
"!For C: x2
4+
y2
9= 1
Use
x = 2cos ! , dx = " 2sin ! d !
y = 3sin ! , dy = 3cos ! d !
!V i d
!r
C
"! = 3sin " (#2sin " ) d " # 2cos " (3cos " ) d " 0
2 $
!0
2 $
!= ! 6 d " = ! 12 #
0
2 #
$
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! "!
V = # 2 k !
V ( x, y) = yi ! x j
For C: x2
4+
y2
9= 1
!V i d
!r
C
"! = " 12 #
Area of ellipse: A = ! ab = 6 !
So,!
V i d !r
C
"! = A i " #!
V
Circulation per unit area
(!
V i d !r )
! A"" = # $
!V ( )
A
"" i kdA
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Example !
V ( x, y) = ! yi
parallel vectors, but circulation non-zero
! "!
V =
k
!V i d
!r
C
"! = " ydx = " sin # (" sin # d # ) = $ 0
2 $
!C
"!
For C: x2
+ y2
= 1
(area)
Again,
!V i d
!r
C
"! = A i " #!
V
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Circulation per unit area (microscopic circulation) in xy plane
= lim! " 0
1
!
!V i d
!r
C
"# = lim! " 0
1
! $ %
!V ( )
!
## i kd !
= ! "!
V ( )i k lim# $ 0
1
# #
%%d # = ! "
!
V ( )i k because in the limit of small
! , " #!
V ( )i k is constant
For a surface element with normal n
lim! " 0
1
!
!V i d
!r
C
"# = $ %!
V i n
d) Stokes theoremConsider any simple curve and a 2-sided surface bounded by it.
(This excludes the Moebius strip: )
For an element of area d !! = d ! n
n
!V i d
!r
! ( d " )"# = $ %
!V i nd "
For a nite surface , adjacent path integrals cancel leaving onlythe outer path:
!V i d
!r
! ( d " )
"## = $ %!
V i n d " "
##
!V i d
!r
! " "# = $ %
!V i
n d " "
## Stokes theorem
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Example !
V = 4 yi + x j + 2 z k
Find ! "!
V i
n d # S
$ for S: x 2 + y 2 + z 2 = a 2 , z ! 0
Stokes law: ! "!
V i n d # S
$ =!
V i d !r for C: x2 + y2 = a 2 in xy plane
C "$
= (4 ydx + x dy )
C
!! = " = " 3# a 2
or , since the integral is the same for any surface bounded by C,
! "!
V i n d # S
$ = ! "!
V i n d # %S $ where %S : is disk x2 + y2 = a 2
= (1 ! 4) k i k d " #S $ =-3 % a 2
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e) Amperes Law
! B i d
!r = 0
! J i d
!! where
! J is the current density through !
!
"C
""
!
C I
! B i d
!r = 0 I
C
"!
By Stokes theorem,
! B i d
!r = ! "
! B i
n d # #
$C
"$so
! "!
B i n d # #
$ = 0!
J i d !
#
#
$
This is true for any so ! "!
B = 0!
J Differential form of
Amperes Law
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f) Conservative elds
For a simply connected region, and F well-behaved,
! "
!
F = 0 # ! "!
F ( )$ %i d
!
$ = 0
!!F i d
!r = 0
C
""
! independence of path
!!
F i d !
r is exact
!!
F = " W
For a eld !
E = !" #
if ! i!
E = 0 (no sources) then ! i ! " = 0
or ! 2 " = 0 Laplaces equation
If ! i!
E = " / # 0
then ! 2 " = # $ / % 0 Poissons equation
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g) Vector Potential
We had, ! "!
V = 0 #!
V = $! % irrotational
If ! i!
V = 0 "!
V = ! #!
A solenoidal
A is the vector potential
Construct A if ! i!
V = 0 such that!
V = ! "!
A
!
V= ! "
!
A represents 3 equations in partial derivatives of A
, but thereare 9 such derivatives: ! A x , y , z! x, y, z
Therefore, there are 6 degrees of freedom.
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Choose A x = 0, using up 3 degrees of freedom.
Then
!
V = ! "!
A =
i
j
k #
# x
#
# y
#
# z A x A y A z
=
i! A z! y
"! A y
! z
#
$ %
&
'("
j ! A z! x
+
k ! A y
! xV
x V y V z
! A y = V z dx + f ( y, z) (1)"Choose f(x,y) = 0, using up 2 degrees of freedom, leaving one
! A z = " V y dx + g( y, z) (2)#
This species A, except for g(y,z), which must be chosen to satisfythe V x equation :
V x =! A z! y
"! A y! z
# $ %
& '(
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V x =! A z! y
"! A y! z
# $ %
& '(
= !"V y" y
dx# + "g( y, z)" y !"V z" z# dx using eq's (1) & (2)
= !"V y" y
+"V z" z
# $ %
& '( dx) + h( y, z)
From
! i!
V = 0,"V x" x
= #"V y" y
#"V z" z
$ % &
' ()
so V x =! V x! x" dx + h( y, z)
= V x ( x, y, z) ! V x (0, y, z) + h( y, z)
! h( y, z) = V x (0, y, z) ="g( y, z)
" y# $ %
& '(
A y = V z dx! A z = ! V y dx + g( y, z)"
A x
= 0
g( y, z) = V x(0, y, z) dy!
Note, A is not unique. ! " ! u = 0 for any u, so
!
! A =!
A + " u also satisfies!
V = " #!
! A
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